Projectile motion (part 3)
An example of solving for the final velocity when you know the change in distance, time, initial velocity, and acceleration. Created by Sal Khan.
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- What is the difference between constant acceleration and constant velocity?(5 votes)
- constant acceleration means the velocity is changing at a constant rate it can increase or decrease but constant velocity means that if a particle is having constant velocity it will move with that velocity only until and unless any external force is applied to it(9 votes)
- Are we supposed to memorize each and every formula for time? I need to know how to get time when given different values ( distance, initial velocity, final velocity, displacement, ect.) Can you give me a list of formulas to use in each scenario? Thanks!!!(4 votes)
- it's quite simple to remember.
where,v is final velocity,u initial velocity,t the time,a acceleration and s the distance.(4 votes)
- Hi Sal,
You have taken upwards positive and downwards negative. Could we take upwards negative and downwards positive?(2 votes)
- Yes, since coordinates have no effect on physics.(4 votes)
- At1:45Salman Khan told that Avg. Velocity = Change in distance by change in time, But I thought that it was Initial vleocity+ Final Velocity /2? Which one is correct?(3 votes)
- Your formula only works when acceleration is constant.(3 votes)
- Isn't sal supposed to use displacement (delta S) instead of distance(1 vote)
- In7:40he stated that the formula for average velocity is (Vf+Vi)/2 assuming the acceleration is constant , but what if the acceleration wasnt constant and we were dealing with a jerky motion what the formula for average velocity might be ?(1 vote)
- Then you would need to calculate the average in a different way, perhaps with calculus.(1 vote)
- My problem has angles and no velocity is given. How do I figure it out when I'm given an angle and how far down the projectile went from the horizon?(1 vote)
- i am unable to do the problems in our text using the formula Sal says?what should i do?(1 vote)
- When we say "change in displacement 'delta S' ", do we mean the change in the displacement form a particular reference point?(3 votes)
- displacement itself is change in position vector. final minus initial. initial can be your reference. i hope you should not say change in displacement.(0 votes)
- Aren't we looking for the time? So why is the last formula is for distance? I tried my best to derive the time from D = ViT + ( ( aT^2 ) / 2 ) but no luck. I was hoping Sal would give the formula for time with the given distance, initial velocity and acceleration. Seems impossible since we need time to determine final velocity.(0 votes)
- transpose everything but time to LHS to find time(2 votes)
In the last video, I said that we started off with the change in distance, so we said that we know the change in distance. These are the things that we are given. We're given the acceleration, we're given the initial velocity, and I asked you how do we figure out what the final velocity is? In the last video-- if you don't remember it, go watch that last video again-- we derived the formula that vf squared, the final velocity squared, is equal to the initial velocity squared plus 2 times the change in distance. You'll sometimes just see it written as 2 times distance, because we assume that the initial distance is at point 0, so the change in distance would just be the final distance. We could write it either way, and hopefully, at this point, you see why I keep switching between change in distance and distance. It's just so you're comfortable when you see it either way. This is for the situation when we didn't know what the vf was. Let's say we want to solve for time instead. Once we solve for the final velocity, we could actually solve for time, and I'll show you how to do that, but let's say we didn't want to go through this step-- how can we solve for time directly, given the change in distance, the acceleration, and the initial velocity? Let's go back once again to the most basic distance formula-- not the distance formula, but how distance relates to velocity. We know that-- I'll write it slightly different this time-- the change in distance over the change in time is equal to the average velocity. We could have rewritten this as the change in distance is equal to the average velocity times the change in time. This is change in time and change in distance. Sometimes we'll just see this written as d equals-- let me write this in a different color, so we have some variety-- velocity times time, or d equals rate times time. The reason why I have change in distance here, or change in time, is that I'm not assuming necessarily that we're starting off at the point 0 or at time 0. If we do, then it just turns out to the final distance is equal to the average velocity times the final time, but let's stick to this. We want to figure out time given this set of inputs. Let's go from this equation. If we want to solve for time, or the change in time, we could just could divide both sides by the average velocity-- actually, no, let's not do that. Let's just stay in terms of change in distance. I've wasted space too fast, so let me clear this and start again. We're given change in distance, initial velocity, and acceleration, and we want to figure out what the time is-- it's really the change in time, but let's just assume that we start time 0, so it's kind of the final time. Let's just start with the simple formula: distance, or change in distance-- I'll use them interchangeably, with a lower case d this time-- is equal to the average velocity times time. What's the average velocity? The average velocity is just the initial velocity plus the final velocity over 2. The only reason why we can just average the initial and the final is because we're assuming constant acceleration, and that's very important, but in most projectile problems, we do have constant acceleration-- downwards-- and that's gravity. We can assume, and we can do this-- we can say that the average of the initial and the final velocity is the average velocity, and then we multiply that times time. Can we use this equation directly? No. we know acceleration, but don't know final velocity. If we can write this final velocity in terms of the other things in this equation, then maybe we can solve for time. Let's try to do that: distance is equal to-- let me take a little side here. What do we know about final velocity? We know that the change in velocity is equal to acceleration times time, assuming that time starts a t equals 0. The change in velocity is the same thing is vf minus vi is equal to acceleration times time. We know that the final velocity is equal to the initial velocity plus acceleration times time. Let's substitute that back into what I was writing right here. We have distance is equal to the initial velocity plus the final velocity, so let's substitute this expression right here. The initial velocity, plus, now the final velocity is now the initial velocity, plus acceleration times time, and then we divide all of that by 2 times time. We get d is equal to-- we have 2 in the numerator, we have 2 initial velocity, 2vi's plus at over 2, and all of that times t. Then we can simplify this. This equals d is equal to-- this 2 cancels out this 2, and then we distribute this t across both terms-- so d is equal to vit plus-- this term is at over 2, but then you multiply the t times here, too-- so it's at squared over 2 plus at squared over 2. We could use this formula if we know the change in distance, or the distance-- this actually should be the change in distance, and the change in time-- is equal to the initial velocity times time plus acceleration times squared divided by 2. Let me summarize all of the equations we have, because we really now have in our arsenal every equation that you really need to solve one dimensional projectile problems-- things going either just left, right, east, west, or north, south, but not both. I will do that in the next video. Let's summarize everything we know. We know the change in distance divided by the change in time is equal to velocity-- average velocity, and it would equal velocity if velocity's not changing, but average when velocity does change-- and we have constant acceleration, which is an important assumption. We know that the change in velocity divided by the change in time is equal to acceleration. We know the average velocity is equal to the final velocity plus the initial velocity over 2, and this assumes acceleration is constant. If we know the initial velocity, acceleration, and the distance, and we want to figure out the final velocity, we could use this formula: vf squared equals vi squared plus 2a times-- really the change in distance, so I'm going to write the change in distance, because that sometimes matters when we're dealing with direction-- change in distance, but so you'll sometimes just write this as distance. Then we just did the equation-- I think I did this in the third video, as well, early on-- but we also learned that distance is equal to the initial velocity times time plus at squared over 2. In that example that I did a couple of videos ago, where we had a cliff-- actually, I only have a minute left in this video. I will do that in the next presentation. I'll see you soon.